What is the solution for y^{''}+3y^'=-e^{2x}+2e^{3x} ?

The solution for y^{''}+3y^'=-e^{2x}+2e^{3x} is y=c_{1}+c_{2}e^{-3x}-1/10 e^{2x}+1/9 e^{3x}

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y^{''}+3y^'=-e^{2x}+2e^{3x}

y^{^{''}} + 3 y^{^{'}} = - e^{2 x} + 2 e^{3 x}

y = c_{1} + c_{2} e^{- 3 x} - \frac{1}{10} e^{2 x} + \frac{1}{9} e^{3 x}

Solution steps

y^{''} (x) + 3 y^{'} (x) = - e^{2 x} + 2 e^{3 x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} - \frac{1}{10} e^{2 x} + \frac{1}{9} e^{3 x}

y = c_{1} + c_{2} e^{- 3 x} - \frac{1}{10} e^{2 x} + \frac{1}{9} e^{3 x}

y^{^{''}} + 25 y = 50 sin (5 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = cos (2 x) - 4 sin (2 x)

y^{^{''}} + 2 y^{^{'}} + y = t^{5} e^{- 5 t}

y^{^{''}} + 16 y = 9 (t) - 9 (t - 98 π), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = 8 e^{- x} + 9

What is the solution for y^{''}+3y^'=-e^{2x}+2e^{3x} ?
The solution for y^{''}+3y^'=-e^{2x}+2e^{3x} is y=c_{1}+c_{2}e^{-3x}-1/10 e^{2x}+1/9 e^{3x}